MULTINOMIAL THEOREM.

(190.) This theorem, as we have just hinted, gives the law of the expansion of

m

(a+a1x+a2x2+a3x3+&c.)",

or of any other polynomial, having for an exponent any whatever. To determine this law, assume

m

(a+a1x+a2x2+&c.)π=A+A1x+A2x2+&c.

value

(1)

m

When x=0, we have an=A, therefore we have

(a+a1x+a2x2+&c.)n=an+A1x+A2x2+&c. Writing x1 for x, we have

(2)

(a+a1x1+a2x}+&c.)ñ=añ+A1x1+А1x}+&c. (3)

Subtracting (3) from (2), we find

m

(a+a1x+a2x2+&c.)ñ−(a+a1x1+a2x}+&c.)ñ

If we suppose

=A1(x−x1)+A2(x2−x})+&c.

U=(a+á1x+a2x2+&c.)»
U2=(a+a1x1+a2x}+&c.) *

m

=(a+a1x+a2x2+&c.)π−(a+a1x1+a2x}+&c.)

U”—U1"=a1(x−x1)+a2(x2−x})+&c.

m

(5)

Dividing the left-hand member of (5) by U-U1", and its right-hand member by its equal.

we get Um-U1

Un-U1

m

=

n

a1(x−x1)+ à2 (x2−x})+&c.,

A1(x−x1)+Á2(x2 −x2)+A3(x3 −→r3)+&c. а 1(x−x1)+α2(x2 −x})+a 3(x3 −x3)+&c.

(6)

If we divide both numerator and denominator of the lefthand member of (6) by U—U1, it will become (see formula

If we divide both numerator and denominator of the righthand member of (6) by x-x1, it will become

A1+A2(x+1)+A3(x2+xx1+x})+&c.
a1+a2(x+x1)+a3(x2+xμμ1+x})+&c.

(8)

The expressions (7) and (S) are equal. Now, when the expression (7) becomes

which, by resubstituting the value of U, becomes

When x=x1, the expression (8) becomes

(9)

1

A1+2A2x+343x2+&c.

Equating the expressions (9) and (10), and clearing of fractions, we have

m

=(a+a1x+a2 x2+&c)TM. (a1+2a2x+3ȧ3x2+&c)= }

m

(a+a1x+a2x2+&c).(A1+2A2x+3A3x2+&c)

(11)

m

m

n=

TM (a+a2x+a2x2+&c.) === (A+A1x+4x2+&c.) (12)

n

Hence (11) becomes

m

n

m (A+A ̧x+A2x2+&c). (a1+2a2x+3a3x2+&c)=

n

(a+a1x+a2x2+&c). (42+2A2x+3A3x2+&c.)

By actual multiplication (13) becomes

=A1a+2A2a\x+3А ̧α\x2+4A1a\x3+&c.

(13)

(14)

Equating coefficients of like powers of x in (14), we have

If for A, we use its equal añ, we shall find from the above system of equations

These values of A, A1, A2, A3, &c., substituted in (1), we have

m

m

m

m

(a+a ̧x+a2x2+azx3 +&c.)n=a+" an

n

Ш

-1

a1a 2+ an

a 3

n

EXAMPLES.

4

1. What is the cube of 1+x+x2+x3 + x2+ &c.?

If, in our general expression (A) of the multinomial theorem, we make a=a1=a2=a3=&c.=1; and m=3, n=1, we shell have

(1+x+x2+x3+x^+&c.)3=1+3x+6x2+10x3+&c.

2. What is the square root of 1+x+x2+x3+&c.? In our general expression (A), we must have m=1, n=2, and 1=a=a1=a2=a3=&c.

(1+x+x2+x3+&c.)2=1+x+_ x2+

1

1 3

3. What is the cube root of 1+x+x2+x3 + &c. ?,

In (A), make m=1, n=3 and 1=a=a1=a2=a3=&C., and we get

1 2

14

(1+x+x2+x3+&c.)3=1+x+x2+

8723

+&c.